Mystery Number 5 Brain Teaser

It’s that time of the week again. Time to tickle your brain. The brain is the center of the nervous system in the human body. Did you know that the brain knows everything but it can’t feel a thing? (there are no pain receptors in the brain). Today’s Brain Teaser is a mathematical one. Some good exercise for your brain.

There is a 10-digit mystery number (no leading 0), represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?

A + B + C + D + E is a multiple of 6.

F + G + H + I + J is a multiple of 5.

A + C + E + G + I is a multiple of 9.

B + D + F + H + J is a multiple of 2.

AB is a multiple of 3.

CD is a multiple of 4.

EF is a multiple of 7.

GH is a multiple of 8.

IJ is a multiple of 10.

FE, HC, and JA are all prime numbers.

Scroll down below the photo to see the answer

The Answer is 5736912480

A = 5, B = 7, C = 3, D = 6, E = 9, F = 1, G = 2, H = 4, I = 8, J = 0

5 + 7 + 3 + 6 + 9 = 30, which is a multiple of 6.

1 + 2 + 4 + 8 + 0 = 15, which is a multiple of 5.

5 + 3 + 9 + 2 + 8 = 27, which is a multiple of 9.

7 + 6 + 1 + 4 + 0 = 18, which is a multiple of 2.

57 is a multiple of 3.

36 is a multiple of 4.

91 is a multiple of 7.

24 is a multiple of 8.

80 is a multiple of 10.

19, 43, and 05 are prime numbers.

A) The sum of digits 0 through 9 is 45. To satisfy Conditions 1 and 2, a pair of multiples must equal 45. The only pair that satisfies those conditions are 30 (multiple of 6) and 15 (multiple of 5). Likewise, to satisfy Conditions 3 and 4, the pair of multiples must be 27 (multiple of 9) and 18 (multiple of 2).

C) To satisfy Condition 9, J must be 0. Delete multiples that require 0 in a different location. For JA to be prime (Condition 10), A must be: 2, 3, 5, or 7. Delete multiples of 3 that have another digit for A.

D) For FE and HC to be prime (Condition 10), C and E can not be even or 5. Delete multiples of 4 and 7 that have another digit for C and E.

E) Per Condition 4, B + D + F + H + J = 18. D, H, and J must be even (required to form multiples of even numbers). To attain the sum of 18 (an even number), B and F must either be both odd or both even. If both are even, then the sum of those five digits would be 20. B and F must be odd. Delete multiples of 3 and 7 that have an even digit for B and F. The only remaining multiple of 7 is 91. E is 9 and F is 1. Delete multiples that require 1 or 9 in a different location.

F) The only remaining options for AB contain 7. Delete multiples that require 7 in a different location. The only remaining options for CD contain 3. Delete multiples that require 3 in a different location.

G) To satisfy Condition 1, CD must be 36 and AB must be either 57 or 75. C is 3 and D is 6. Delete multiples that require 5, 6, or 7 in a different location.

H) The only remaining options for GH contain 4. Delete multiples that require 4 in a different location.

I) Per Condition 3, A + C + E + G + I = 27. Since C =3 and I = 9, the other three must total 15. There is only one combination of remaining multiples (57, 24, and 80) that can satisfy this condition. A is 5, G is 2, and I is 8. B is 7 and H is 4.

Sherbet, spent half of my working day wrecking my brain!!!!! but was forced to scroll down avoiding 2 days of getting the answer myself..lol – I have to admit its a real teazer & definately brain fodder